JEE Challenger
More from Mechanical Properties of Fluids
PhyHardAdv 2021 • Paper 1

Pressure Difference in Accelerated Cylindrical Tube with Water

Consider a cylindrical vessel filled with water, positioned as illustrated in the figure. It moves downward with a uniform acceleration aa along a fixed inclined plane angled at θ=45∘\theta = 45^\circ. The pressures at points 1 and 2 located at the bottom of the tube are P1P_1 and P2P_2, respectively. Let β=P1−P2ρgd\beta = \frac{P_1 - P_2}{\rho g d}, where ρ\rho denotes the fluid density, dd is the inner diameter of the tube, and gg is the acceleration due to gravity. Which of the following option(s) is(are) correct?

Question Diagram 1

Options

A

β=0\beta = 0 when a=g/2a = g/\sqrt{2}

Correct
B

β>0\beta > 0 when a=g/2a = g/\sqrt{2}

C

β=2−12\beta = \frac{\sqrt{2}-1}{\sqrt{2}} when a=g/2a = g/2

Correct
D

β=12\beta = \frac{1}{\sqrt{2}} when a=g/2a = g/2

Topics & Concepts

Mechanical Properties of FluidsFluid Dynamics

Step-by-Step Solution

To determine the pressure difference between points 11 and 22 located at the bottom of the cylindrical vessel, we analyze the fluid dynamics in the non-inertial frame of reference attached to the accelerating vessel.

1. Frame of Reference and Effective Acceleration

The vessel moves down the inclined plane (angled at θ=45∘\theta = 45^\circ) with a uniform acceleration aa. Defining a Cartesian coordinate system with i^\hat{i} pointing horizontally to the right and j^\hat{j} pointing vertically upwards:

  • The acceleration vector a⃗\vec{a} of the vessel is given by: a⃗=acos⁡45∘i^−asin⁡45∘j^=a2i^−a2j^\vec{a} = a \cos 45^\circ \hat{i} - a \sin 45^\circ \hat{j} = \frac{a}{\sqrt{2}} \hat{i} - \frac{a}{\sqrt{2}} \hat{j}

  • The acceleration due to gravity g⃗\vec{g} is: g⃗=−gj^\vec{g} = -g \hat{j}

In the non-inertial reference frame of the vessel, the effective acceleration due to gravity g⃗eff\vec{g}_{\text{eff}} acting on the fluid is: g⃗eff=g⃗−a⃗=−a2i^−(g−a2)j^\vec{g}_{\text{eff}} = \vec{g} - \vec{a} = - \frac{a}{\sqrt{2}} \hat{i} - \left(g - \frac{a}{\sqrt{2}}\right) \hat{j}

2. Pressure Gradient and Difference

The pressure gradient inside the accelerated fluid is governed by: ∇P=ρg⃗eff=−ρa2i^−ρ(g−a2)j^\nabla P = \rho \vec{g}_{\text{eff}} = -\frac{\rho a}{\sqrt{2}} \hat{i} - \rho\left(g - \frac{a}{\sqrt{2}}\right) \hat{j}

Let r⃗1\vec{r}_1 and r⃗2\vec{r}_2 be the position vectors of points 11 and 22, respectively. Since the base of the tube lies along the inclined plane at an angle θ=45∘\theta = 45^\circ:

  • The horizontal separation between points 11 and 22 is dd.
  • Point 11 is lower than point 22 along the vertical by dtan⁡45∘=dd \tan 45^\circ = d.

Thus, the displacement vector from point 22 to point 11 is: Δr⃗=r⃗1−r⃗2=di^−dj^\Delta \vec{r} = \vec{r}_1 - \vec{r}_2 = d \hat{i} - d \hat{j}

The pressure difference between points 11 and 22 is: P1−P2=∇P⋅Δr⃗P_1 - P_2 = \nabla P \cdot \Delta \vec{r} P1−P2=(−ρa2)(d)+(−ρ(g−a2))(−d)P_1 - P_2 = \left(-\frac{\rho a}{\sqrt{2}}\right)(d) + \left(-\rho\left(g - \frac{a}{\sqrt{2}}\right)\right)(-d) P1−P2=−ρad2+ρgd−ρad2=ρgd−2ρad=ρd(g−2a)P_1 - P_2 = -\frac{\rho a d}{\sqrt{2}} + \rho g d - \frac{\rho a d}{\sqrt{2}} = \rho g d - \sqrt{2}\rho a d = \rho d \left(g - \sqrt{2}a\right)

3. Evaluating β\beta

The dimensionless parameter β\beta is defined as: β=P1−P2ρgd=ρd(g−2a)ρgd=1−2ag\beta = \frac{P_1 - P_2}{\rho g d} = \frac{\rho d (g - \sqrt{2}a)}{\rho g d} = 1 - \frac{\sqrt{2}a}{g}

Now we test the given conditions:

  1. For a=g2a = \frac{g}{\sqrt{2}}: β=1−2g(g2)=1−1=0\beta = 1 - \frac{\sqrt{2}}{g} \left(\frac{g}{\sqrt{2}}\right) = 1 - 1 = 0 Therefore, (A) is correct and (B) is incorrect.

  2. For a=g2a = \frac{g}{2}: β=1−2g(g2)=1−12=2−12\beta = 1 - \frac{\sqrt{2}}{g} \left(\frac{g}{2}\right) = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}} Therefore, (C) is correct and (D) is incorrect.

Conclusion

The correct options are A and C.

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